| // Copyright 2023 Google LLC | |
| // Use of this source code is governed by a BSD-style license that can be found in the LICENSE file. | |
| #include "modules/bentleyottmann/include/Segment.h" | |
| #include "include/private/SkAssert.h" | |
| #include "include/private/SkTo.h" | |
| #include "modules/bentleyottmann/include/Int96.h" | |
| #include <algorithm> | |
| #include <cmath> | |
| namespace bentleyottmann { | |
| // -- Segment -------------------------------------------------------------------------------------- | |
| Point Segment::upper() const { | |
| return std::min(p0, p1); | |
| } | |
| Point Segment::lower() const { | |
| return std::max(p0, p1); | |
| } | |
| // Use auto [l, t, r, b] = s.bounds(); | |
| std::tuple<int32_t, int32_t, int32_t, int32_t> Segment::bounds() const { | |
| auto [l, r] = std::minmax(p0.x, p1.x); | |
| auto [t, b] = std::minmax(p0.y, p1.y); | |
| return std::make_tuple(l, t, r, b); | |
| } | |
| bool operator==(const Segment& s0, const Segment& s1) { | |
| return s0.upper() == s1.upper() && s0.lower() == s1.lower(); | |
| } | |
| bool operator<(const Segment& s0, const Segment& s1) { | |
| return std::make_tuple(s0.upper(), s0.lower()) < std::make_tuple(s1.upper(), s1.lower()); | |
| } | |
| bool no_intersection_by_bounding_box(const Segment& s0, const Segment& s1) { | |
| auto [left0, top0, right0, bottom0] = s0.bounds(); | |
| auto [left1, top1, right1, bottom1] = s1.bounds(); | |
| // If the sides of the box touch, then there is no new intersection. | |
| return right0 <= left1 || right1 <= left0 || bottom0 <= top1 || bottom1 <= top0; | |
| } | |
| // Derivation of Intersection | |
| // The intersection point I = (X, Y) of the two segments (x0, y0) -> (x1, y1) | |
| // and (x2, y2) -> (x3, y3). | |
| // X = x0 + s(x1 - x0) = x2 + t(x3 - x2) | |
| // Y = y0 + s(y1 - y0) = y2 + t(y3 - y2) | |
| // | |
| // Solve for s in terms of x. | |
| // x0 + s(x1 - x0) = x2 + t(x3 - x2) | |
| // s(x1 - x0) = x2 - x0 + t(x3 - x2) | |
| // s = (x2 - x0 + t(x3 - x2)) / (x1 - x0) | |
| // | |
| // Back substitute s into the equation for Y. | |
| // y0 + ((x2 - x0 + t(x3 - x2)) / (x1 - x0))(y1 - y0) = y2 + t(y3 - y2) | |
| // (x2 - x0 + t(x3 - x2)) / (x1 - x0) = (y2 - y0 + t(y3 - y2)) / (y1 - y0) | |
| // (y1 - y0)(x2 - x0 + t(x3 - x2)) = (x1 - x0)(y2 - y0 + t(y3 - y2)) | |
| // (y1 - y0)(x2 - x0) + t(y1 - y0)(x3 - x2) = (x1 - x0)(y2 - y0) + t(x1 - x0)(y3 - y2) | |
| // Collecting t's on one side, and constants on the other. | |
| // t((y1 - y0)(x3 - x2) - (x1 - x0)(y3 - y2)) = (x1 - x0)(y2 - y0) - (y1 - y0)(x2 - x0) | |
| // | |
| // Solve for t in terms of x. | |
| // x0 + s(x1 - x0) = x2 + t(x3 - x2) | |
| // x0 - x2 + s(x1 - x0) = t(x3 - x2) | |
| // (x0 - x2 + s(x1 - x0)) / (x3 - x2) = t | |
| // Back substitute t into the equation for Y. | |
| // y0 + s(y1 - y0) = y2 + ((x0 - x2 + s(x1 - x0)) / (x3 - x2))(y3 - y2) | |
| // (y0 - y2 + s(y1 - y0)) / (y3 - y2) = (x0 - x2 + s(x1 - x0)) / (x3 - x2) | |
| // (x3 - x2)(y0 - y2 + s(y1 - y0)) = (y3 - y2)(x0 - x2 + s(x1 - x0)) | |
| // (x3 - x2)(y0 - y2) + s(x3 - x2)(y1 - y0) = (y3 - y2)(x0 - x2) + s(y3 - y2)(x1 - x0) | |
| // Collecting s's on on side and constants on the other. | |
| // s((x3 - x2)(y1 - y0) - (y3 - y2)(x1 - x0)) = (y3 - y2)(x0 - x2) - (x3 - x2)(y0 - y2) | |
| // Assign names and vectors to extract the cross products. The vector (x0, y0) -> (x1, y1) is | |
| // P0 -> P1, and is named Q = (x1 - x0, y1 - y0) = P1 - P0. The following vectors are defined in | |
| // a similar way. | |
| // * Q: P1 - P0 | |
| // * R: P2 - P0 | |
| // * T: P3 - P2 | |
| // Extracting cross products from above for t. | |
| // t((P3 - P2) x (P1 - P0)) = (P1 - P0) x (P2 - P0) | |
| // t(T x Q) = Q x R | |
| // t = (Q x R) / (T x Q) | |
| // Extracting cross products from above for t. | |
| // s((P3 - P2) x (P1 - P0)) = (P0 - P2) x (P3 - P2) | |
| // s(T x Q) = -R x T | |
| // s = (T x R) / (T x Q) | |
| // | |
| // There is an intersection only if t and s are on [0, 1]. | |
| // | |
| // This method of calculating the intersection only uses 8 multiplies, and 1 division. It also | |
| // determines if the two segments cross with no round-off error and is always correct using 6 | |
| // multiplies. However, the actual crossing point is rounded to fit back into the int32_t. | |
| std::optional<Point> intersect(const Segment& s0, const Segment& s1) { | |
| // Check if the bounds intersect. | |
| if (no_intersection_by_bounding_box(s0, s1)) { | |
| return std::nullopt; | |
| } | |
| // Create the end Points for s0 and s1 | |
| const Point P0 = s0.upper(), | |
| P1 = s0.lower(), | |
| P2 = s1.upper(), | |
| P3 = s1.lower(); | |
| if (P0 == P2 || P1 == P3 || P1 == P2 || P3 == P0) { | |
| // Lines don't intersect if they share an end point. | |
| return std::nullopt; | |
| } | |
| // Create the Q, R, and T. | |
| const Point Q = P1 - P0, | |
| R = P2 - P0, | |
| T = P3 - P2; | |
| // 64-bit cross product. | |
| auto cross = [](const Point& v0, const Point& v1) { | |
| int64_t x0 = SkToS64(v0.x), | |
| y0 = SkToS64(v0.y), | |
| x1 = SkToS64(v1.x), | |
| y1 = SkToS64(v1.y); | |
| return x0 * y1 - y0 * x1; | |
| }; | |
| // Calculate the cross products needed for calculating s and t. | |
| const int64_t QxR = cross(Q, R), | |
| TxR = cross(T, R), | |
| TxQ = cross(T, Q); | |
| if (TxQ == 0) { | |
| // Both t and s are either < 0 or > 1 because the denominator is 0. | |
| return std::nullopt; | |
| } | |
| // t = (Q x R) / (T x Q). s = (T x R) / (T x Q). Check that t & s are on [0, 1] | |
| if ((QxR ^ TxQ) < 0 || (TxR ^ TxQ) < 0) { | |
| // The division is negative and t or s < 0. | |
| return std::nullopt; | |
| } | |
| if (TxQ > 0) { | |
| if (QxR > TxQ || TxR > TxQ) { | |
| // t or s is greater than 1. | |
| return std::nullopt; | |
| } | |
| } else { | |
| if (QxR < TxQ || TxR < TxQ) { | |
| // t or s is greater than 1. | |
| return std::nullopt; | |
| } | |
| } | |
| // Calculate the intersection using doubles. | |
| // TODO: This is just a placeholder approximation for calculating x and y should use big math | |
| // above. | |
| const double t = static_cast<double>(QxR) / static_cast<double>(TxQ); | |
| SkASSERT(0 <= t && t <= 1); | |
| const int32_t x = std::round(t * (P3.x - P2.x) + P2.x), | |
| y = std::round(t * (P3.y - P2.y) + P2.y); | |
| return Point{x, y}; | |
| } | |
| // The comparison is: | |
| // x0 + (y - y0)(x1 - x0) / (y1 - y0) <? x2 + (y - y2)(x3 - x2) / (y3 - y2) | |
| // Factor out numerators: | |
| // [x0(y1 - y0) + (y - y0)(x1 - x0)] / (y1 - y0) <? [x2(y3 - y2) + (y - y2)(x3 -x 2)] / (y3 - y2) | |
| // Removing the divides by cross multiplying. | |
| // [x0(y1 - y0) + (y - y0)(x1 - x0)] (y3 - y2) <? [x2(y3 - y2) + (y - y2)(x3 - x2)] (y1 - y0) | |
| // This is a 64-bit int x0 + (y - y0) (x1 - x0) times a 32-int (y3 - y2) resulting in a 96-bit int, | |
| // and the same applies to the other side of the <?. Because y0 <= y1 and y2 <= y3, then the | |
| // differences of (y1 - y0) and (y3 - y2) are positive allowing us to multiply through without | |
| // worrying about sign changes. | |
| bool less_than_at(const Segment& s0, const Segment& s1, int32_t y) { | |
| auto [l0, t0, r0, b0] = s0.bounds(); | |
| auto [l1, t1, r1, b1] = s1.bounds(); | |
| SkASSERT(t0 <= y && y <= b0); | |
| SkASSERT(t1 <= y && y <= b1); | |
| // Return true if the bounding box of s0 is fully to the left of s1. | |
| if (r0 < l1) { | |
| return true; | |
| } | |
| // Return false if the bounding box of s0 is fully to the right of s1. | |
| if (r1 < l0) { | |
| return false; | |
| } | |
| // Check the x intercepts along the horizontal line at y. | |
| // Make s0 be (x0, y0) -> (x1, y1) and s1 be (x2, y2) -> (x3, y3). | |
| auto [x0, y0] = s0.upper(); | |
| auto [x1, y1] = s0.lower(); | |
| auto [x2, y2] = s1.upper(); | |
| auto [x3, y3] = s1.lower(); | |
| int64_t s0YDiff = y - y0, | |
| s1YDiff = y - y2, | |
| s0YDelta = y1 - y0, | |
| s1YDelta = y3 - y2, | |
| x0Offset = x0 * s0YDelta + s0YDiff * (x1 - x0), | |
| x2Offset = x2 * s1YDelta + s1YDiff * (x3 - x2); | |
| Int96 s0Factor = multiply(x0Offset, y3 - y2), | |
| s1Factor = multiply(x2Offset, y1 - y0); | |
| return s0Factor < s1Factor; | |
| } | |
| bool point_less_than_segment_in_x(Point p, const Segment& segment) { | |
| auto [l, t, r, b] = segment.bounds(); | |
| // Ensure that the segment intersects the horizontal sweep line | |
| SkASSERT(t <= p.y && p.y <= b); | |
| // Fast answers using bounding boxes. | |
| if (p.x < l) { | |
| return true; | |
| } else if (p.x >= r) { | |
| return false; | |
| } | |
| auto [x0, y0] = segment.upper(); | |
| auto [x1, y1] = segment.lower(); | |
| auto [x2, y2] = p; | |
| // For a point and a segment the comparison is: | |
| // x2 < x0 + (y2 - y0)(x1 - x0) / (y1 - y0) | |
| // becomes | |
| // (x2 - x0)(y1 - y0) < (x1 - x0)(y2 - y0) | |
| // We don't need to worry about the signs changing in the cross multiply because (y1 - y0) is | |
| // always positive. Manipulating a little further derives predicate 2 from "Robust Plane | |
| // Sweep for Intersecting Segments" page 9. | |
| // 0 < (x1 - x0)(y2 - y0) - (x2 - x0)(y1 - y0) | |
| // becomes | |
| // | x1-x0 x2-x0 | | |
| // 0 < | y1-y0 y2-y0 | | |
| return SkToS64(x2 - x0) * SkToS64(y1 - y0) < SkToS64(y2 - y0) * SkToS64(x1 - x0); | |
| } | |
| // The design of this function allows its use with std::lower_bound. lower_bound returns the | |
| // iterator to the first segment where rounded_point_less_than_segment_in_x_lower returns false. | |
| // Therefore, we want s(y) < (x - ½) to return true, then start returning false when s(y) ≥ (x - ½). | |
| bool rounded_point_less_than_segment_in_x_lower(const Segment& s, Point p) { | |
| const auto [l, t, r, b] = s.bounds(); | |
| const auto [x, y] = p; | |
| // Ensure that the segment intersects the horizontal sweep line | |
| SkASSERT(t <= y && y <= b); | |
| // In the comparisons below, x is really x - ½ | |
| if (r < x) { | |
| // s is entirely < p. | |
| return true; | |
| } else if (x <= l) { | |
| // s is entirely > p. This also handles vertical lines, so we don't have to handle them | |
| // below. | |
| return false; | |
| } | |
| const auto [x0, y0] = s.upper(); | |
| const auto [x1, y1] = s.lower(); | |
| // Horizontal - from the guards above we know that p is on s. | |
| if (y0 == y1) { | |
| return false; | |
| } | |
| // s is not horizontal or vertical. | |
| SkASSERT(x0 != x1 && y0 != y1); | |
| // Given the segment upper = (x0, y0) and lower = (x1, y1) | |
| // x0 + (x1 - x0)(y - y0) / (y1 - y0) < x - ½ | |
| // (x1 - x0)(y - y0) / (y1 - y0) < x - x0 - ½ | |
| // Because (y1 - y0) is always positive we can multiply through the inequality without | |
| // worrying about sign changes. | |
| // (x1 - x0)(y - y0) < (x - x0 - ½)(y1 - y0) | |
| // (x1 - x0)(y - y0) < ½(2x - 2x0 - 1)(y1 - y0) | |
| // 2(x1 - x0)(y - y0) < (2(x - x0) - 1)(y1 - y0) | |
| return 2 * SkToS64(x1 - x0) * SkToS64(y - y0) < (2 * SkToS64(x - x0) - 1) * SkToS64(y1 - y0); | |
| } | |
| // The design of this function allows use with std::lower_bound. lower_bound returns the iterator | |
| // to the first segment where rounded_point_less_than_segment_in_x_upper is false. This function | |
| // implements s(y) < (x + ½). | |
| bool rounded_point_less_than_segment_in_x_upper(const Segment& s, Point p) { | |
| const auto [l, t, r, b] = s.bounds(); | |
| const auto [x, y] = p; | |
| // Ensure that the segment intersects the horizontal sweep line | |
| SkASSERT(t <= y && y <= b); | |
| // In the comparisons below, x is really x + ½ | |
| if (r <= x) { | |
| // s is entirely < p. | |
| return true; | |
| } else if (x < l) { | |
| // s is entirely > p. This also handles vertical lines, so we don't have to handle them | |
| // below. | |
| return false; | |
| } | |
| const auto [x0, y0] = s.upper(); | |
| const auto [x1, y1] = s.lower(); | |
| // Horizontal - from the guards above we know that p is on s. | |
| if (y0 == y1) { | |
| return false; | |
| } | |
| // s is not horizontal or vertical. | |
| SkASSERT(x0 != x1 && y0 != y1); | |
| // Given the segment upper = (x0, y0) and lower = (x1, y1) | |
| // x0 + (x1 - x0)(y - y0) / (y1 - y0) < x + ½ | |
| // (x1 - x0)(y - y0) / (y1 - y0) < x - x0 + ½ | |
| // Because (y1 - y0) is always positive we can multiply through the inequality without | |
| // worrying about sign changes. | |
| // (x1 - x0)(y - y0) < (x - x0 + ½)(y1 - y0) | |
| // (x1 - x0)(y - y0) < ½(2x - 2x0 + 1)(y1 - y0) | |
| // 2(x1 - x0)(y - y0) < (2(x - x0) + 1)(y1 - y0) | |
| return 2 * SkToS64(x1 - x0) * SkToS64(y - y0) < (2 * SkToS64(x - x0) + 1) * SkToS64(y1 - y0); | |
| } | |
| int compare_slopes(const Segment& s0, const Segment& s1) { | |
| Point s0Delta = s0.lower() - s0.upper(), | |
| s1Delta = s1.lower() - s1.upper(); | |
| // Handle the horizontal cases to avoid dealing with infinities. | |
| if (s0Delta.y == 0 || s1Delta.y == 0) { | |
| if (s0Delta.y != 0) { | |
| return -1; | |
| } else if (s1Delta.y != 0) { | |
| return 1; | |
| } else { | |
| return 0; | |
| } | |
| } | |
| // Compare s0Delta.x / s0Delta.y ? s1Delta.x / s1Delta.y. I used the alternate slope form for | |
| // two reasons. | |
| // * no change of sign - since the delta ys are always positive, then I don't need to worry | |
| // about the change in sign with the cross-multiply. | |
| // * proper slope ordering - the slope monotonically increases from the smallest along the | |
| // negative x-axis increasing counterclockwise to the largest along | |
| // the positive x-axis. | |
| int64_t lhs = SkToS64(s0Delta.x) * SkToS64(s1Delta.y), | |
| rhs = SkToS64(s1Delta.x) * SkToS64(s0Delta.y); | |
| if (lhs < rhs) { | |
| return -1; | |
| } else if (lhs > rhs) { | |
| return 1; | |
| } else { | |
| return 0; | |
| } | |
| } | |
| } // namespace bentleyottmann |